This is way outside my area of expertise, so perhaps someone can explain why the answer does not follow from the double suspension theorem: start with the Poincare dodecahedral space $M$ (a homology 3-sphere with nontrivial fundamenatal group) and suspend it once. If you get something homeomorphic to $S^4$, then $M$ is a counterexample. If not, then by DST $SM$ is homeomorphic to $S^5$ so $SM$ is a counterexample.
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Pete L. ClarkMay 5 '11 at 18:29

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Interesting plan! But it is not obvious to me that the suspension of M (or any other space obtained by a similar method) is a topological manifold.
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James CranchMay 5 '11 at 18:54

Okay, so that's what I was missing: that the suspension of a manifold might or might not be a manifold. Like I said: not my area of expertise. (I guess the upvotes on my previous comment mean: "yes, I was wondering that too...")
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Pete L. ClarkMay 5 '11 at 20:15

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Yes, it's pretty easy that the suspension of a space $X$ cannot possibly be an $n+1$-manifold unless $X$ is homotopy equivalent to $S^n$.
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Tom GoodwillieMay 6 '11 at 0:37

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Suppose $M$ is a closed $n$-manifold whose suspension is homeomorphic to $S^{n+1}$.
Removing the two "singular" points from the suspension gives $M\times \mathbb R$, while
removing two points from $S^{n+1}$ gives $S^n\times\mathbb R$. Thus $M\times \mathbb R$ and $S^n\times\mathbb R$ are homeomorphic, which easily implies that
$M$ and $S^n$ are h-cobordant, and hence $M$ and $S^n$ are homeomorphic.

Topological h-cobordism theorem for simply-connected manifolds holds in all dimensions (due to Freedman in dimension 4, to Perelman in dimension 3, and to Newman in dimensions >4).
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Igor BelegradekMay 5 '11 at 19:15

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To see that $M$ and $S^n$ are h-cobordant consider a homeomorphism $h$ of their products with $\mathbb R$, and use excision in homology to show that the submanifolds $S^n\times 0$ and $h(M\times t)$ bound an h-cobordism, where $t$ need to be sufficiently large to ensure that the submanifolds are disjoint.
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Igor BelegradekMay 5 '11 at 19:38

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In fact, one need not involve h-cobordisms at all: just note that $M$ and $S^n$ are homotopy equivalent and use Poincare's conjecture. I guess, I just like to advertize that fact that if two closed manifolds become homeomorphic after multiplying by $\mathbb R$, then they are $h$-cobordant. :)
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Igor BelegradekMay 5 '11 at 20:11

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@Willie: but then what do we call it? It's not any one person's theorem...
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Pete L. ClarkMay 5 '11 at 21:25